How do you find the antiderivative of $e^(2x)*sin(e^x)dx$ ?

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Answer 1 · 2017-01-17T15:57:16

$inte^(2x)sin(e^x)dx=-e^xcos(e^x)+sin(e^x)+C$

$I=inte^(2x)sin(e^x)dx$

Let $t=e^x$ . This implies that $dt=e^xdx$ . Before substituting, note that $e^(2x)=e^x(e^x)$ .

$I=inte^xsin(e^x)(e^xdx)=inttsin(t)dt$

To do this, use integration by parts. This takes the form $intudv=uv-intvdu$ . Let:

${(u=t" "=>" "du=dt),(dv=sin(t)dt" "=>" "v=-cos(t)):}$

Then:

$I=uv-intvdu=-tcos(t)-int(-cos(t))dt$

$I=-tcos(t)+intcos(t)dt=-tcos(t)+sin(t)$

Returning to the original variable using $t=e^x$ :

$I=-e^xcos(e^x)+sin(e^x)+C$

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